The centripetal force of a child riding a merry-go-round
Problem:
(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?
(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?
(c) Compare each force with her weight.
Solution:
Part A
We are given the following values: m=22.0 kg, ω=40.0 rev/min, and r=1.25 m. We are asked to solve for the centripetal force, Fc.
Centripetal force Fc is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v and has magnitude Fc=mac, which can also be expressed as Fc=mrv2 or Fc=mrω2. Basing from the given values, we are going to solve the problem using the formula
Fc=mrω2
First, we need to convert the angular velocity ω to rad/sec for unit homogeneity.
40 rev/min×1 rev2π rad×60 sec1 min=4.1888 rad/sec
Now, we can substitute the given values into our formula.
FcFcFcFc=mrω2=(22.0 kg)(1.25 m)(4.1888 rad/sec)2=482.5162 N=483 N (Answer)
Part B
Let us convert the angular velocity to radians per second.
3.00 rev/min×1 rev2π rad×60 sec1 min=0.3142 rad/sec
Now, we can substitute the given values
FcFcFcFc=mrω2=(22.0 kg)(8.00 m)(0.3142 rad/sec)2=17.3750 N=17.4 N (Answer)
Part C
For the first centripetal force we solved in Part A,
wFc=(22 kg)(9.81 m/s2)483 N=2.24
The centripetal force is 2.24 times the weight of the child.
For the centripetal force we solved in Part B, we have
wFc=(22 kg)(9.81 m/s2)17.4 N=0.0806
The centripetal force is only about 8% of the child’s weight.
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